Organizing and Displaying Categorical Data Guide Part 7

HW6#16 In the last quarter of 2007, a group of 64 mutual funds had a mean return of 4.4% with a standard deviation of 6.3%. If a normal model can be used to model them, what percent of the funds would you expect to be in each region? Use the 68-95-99.7 rule to approximate the probabilities rather than using technology to find the values more precisely. Be sure to draw a picture first. a. What percent of funds have a return of less than or equal to -8.2%? ANS: 2.5% b. What percent of funds have a return between -14.5% and 23.3% i. .9985-0.0015= .997 = 99.7% c. What percent of funds have a return greater than 10.7%? i. 1-0.84=0.16 ii. Approximate answers using this rule is 0.16 Note that ● Z-score outcomes = outcome’s distance from the mean in S.D. units ● Percentile = probability less than d. How many SD above or below the mean is outcome 17%? i. 2 SD Above e. Base on the picture above what percent of the time does a hedge fund return less than 8% (Trick) i. We don’t know ii. Somewhere between 50% and 85%. Maybe 66.85% or 65%? iii. Z-score of 8= (8-mean)/sd = (8-4.4)/6.3=0.5714 iv. This z-score of 8 is not a whole number. The 68-95-99.7 rule only works for z-scores that are natural numbers, such as 1 or -1. TO ANSWER THIS QUESTION WE USE EXCEL =norm.dist(8,4.4,6.3,1) =norm.dist(x,mean,sd,1) 1= Less than 8% 0= Exactly 8% 1. Using Excel, find the prob the annual return for the group of funds is less than 3? a. =NORM.DIST(3,4.4,6.3,1) = 0.412070448 2. Using Excel, find the prob the annual return for the group of funds is less than -1? a. =NORM.DIST(-1,4.4,6.3,1) = 0.195682969 3. Find the probability the annual return for the group is greater than >3? a. 1-NORM.DIST(3,4.4,6.3,1) 1-0.412070448= 0.587929552 4. Find the probability the annual return for the group is greater than -1? a. 1-NORM.DIST(-1,4.4,6.3,1) 1-0.195682969= 0.804317031 5. Find the probability the annual return for the group is between -1 and 3? a. 0.412-0.196= 0.216 1. What percent of the time do funds return less than 3? Step 1: What is the z-score of 3? Z-score of 3= (3-4.4)/6.3 = -0.222 Step 2: Using the Z-score and excel, find the percentile? =norm.dist(-.222,0,1,1)= 0.412156942 2. What percent of the time do funds return less than -1? Step 1: What is the z-score of -1 Z-score of -1= (-1-4.4)/6.3 = -0.857142857 Step 2: Using the Z-score and excel, find the percentile? =norm.dist(-0.857,0,1,1)= 0.195682969 HW6#6 (Z-Score Norm.Dist) Find the probability Z is less than 0.5 =norm.dist(0.5,0,1,1)= 0.691462461 Find the probability Z is between 0.5 and 0.95 =.83-.69=0.14 Using Excel, find the probability Z is between -1.3 and 1.9? =norm.dist(1.9,0,1,1)-norm.dist(-1.3,0,1,1)= 0.874482956 Find the probability the absolute value of Z is greater than 0.7? =norm.dist(0.7,0,1,1)=0.76 =norm.dist(-0.7,0,1,1)=0.24 Find the probability that z is greater than +0.7 or less than -0.7= absolute Z being greeted than 0.7 ANSWER= 0.24+0.24= 0.48 0.24+(1-0.76) Percentile to Z-Score of Outcome Questions on Midterm Exam HW6 #7,18,19,23 HW6 Question 7 In a standard Normal model, what value(s) of z cut(s) off the region described? Don't forget to draw a picture. a) the highest 25% In norm.dist, we are saying if someone has a score of 30, find his rank. In norm.inv, we are saying if someone ranks 0.75, find their score. What z-score outcome is associated with a percentile of 0.75? =norm.inv(0.75,0,1) =0.6745 z-score - If the percentile is 0.75 and the distribution has a mean of 0 and a SD of 1, find the outcome associated with it. b) the highest 97% What z-score outcome is associated with the 0.03 percentile? =norm.inv(0.03,0,1)=-1.8808 c) the lowest 80% Find the z-score outcome associated with the 80th percentile =norm.inv(0.8,0,1)=0.841621234 d) the middle 70% Percentile 1= 15th Percentile 2=85th The z-scores that cut off the middle 70 % are =norm.inv(0.15,0,1)= -1.0364 =norm.inv(0.85,0,1)= 1.0364 HW6 Question 18 The daily exchange rates for the five-year period 2003 to 2008 between currency A and currency B are well modeled by a normal distribution with mean 1.388 in currency A (to currency B) and standard deviation 0.015 in currency A. Given this model, and using the 68-95-99.7 rule to approximate the probabilities rather than using technology to find the values more precisely, complete parts (a) through (d). a) What would the cutoff rate be that would separate the highest 2.5% of currency A/currency B rates? - Outcome = 1.418 b) What would the cutoff rate be that would separate the lowest 50% of currency A/currency B rates? - Outcome = 1.388 c) The upper cutoff outcome and the lowest cutoff outcome for the middle 68% - Middle 68%= 1SD - Lower cutoff= 1.373; The upper cutoff is 1.403 d) The cutoff for the highest 16% - 1.403 HW6 Question 19 Mean = 95 SD = 13 a. If someone ranks 0.12, what is his score/ outcome? i. =norm.inv(0.12,95,13) b. Find the 90th percentile i. =norm.inv(0.9,95,13) c. The outcome associated with the 25th percentile =norm.inv(0.25,95,13) The outcome associated with the 75th percentile =norm.inv(0.75,95,13) - IQR= 75th-25th = 17.54 HW6 Question 23 A certain tennis player makes a successful first serve 75% of the time. Suppose the tennis player serves 80 times in a match. What's the probability that she makes at least 66 first serves? When she serves one time, the serve is either in or out. The distribution is Bernoulli. When she serves 80 times, T=total # of serves in is distributed Binomial Find the probability that T is less than or equal to 66 using excel 11/8/2020 Distribution of Binomial & Proportion HW7#10 The proportion of adult women in a certain geographical region is approximately 48%. A marketing survey telephones 100 people at random. Complete parts a and b below. a) Sample proportion = 0.48 b) Reach 100*0.48= 48 women HW7#11 An investment website can tell what devices are used to access the site. The site managers wonder whether they should enhance the facilities for trading via "smart phones", so they want to estimate the proportion of users who access the site that way (even if they also use their computers sometimes). They draw a random sample of 300 investors from their customers. Suppose that the true proportion of smartphone users is 43%. a) What is the expected proportion of persons in your sample who have a smartphone? =43% b) What would the standard deviation of the sampling distribution of the proportion of the smartphone users be? Sample SD= SQRT((p)(1-p)/n) = SQRT ((.43)(.57)/300)) =0.028583212 HW7#11 Suppose 29% of the population are smartphone users and we sample 300 people. a) What is the expected sample proportion? i) 0.29 b) Will the sample proportion always be 0.29? i) Will we always get 87 people with smartphones in our sample? NO ii) Because of SAMPLING VARIATION, the sample proportion will not always be 0.29. On average, the sample proportion will be 0.29 c) Using formula, what is the SD of our sample proportion? i) SD(P)= SQRT ((p(1-p)/n) =0.026197964 d) If we pretend the sample proportion is normally distributed, what is the distribution of the sample proportion? P~N(0.29,0.026) The sample proportion is distributed normal with a mean of 0.29 and a SD of 0.026 Note: THE SAMPLE PROPORTION IS NOT A BINOMIAL DISTRIBUTION. The sample proportion relative freq is approx normal e) What is the probability the sample proportion is less than 0.29? i) 0.5 or Norm.dist(.29,.29,.026,1) f) What is the probability the sample proportion is less than 0.27? i) 0.22 or Norm.dist(.27,.29,.026,1) g) What is the probability the sample proportion is greater than 0.27? i) 0.78 or 1-Norm.dist(.27,.29,.026,1) h) What is the probability the sample proportion is less than 0.23? i) 0.011 or Norm.dist(.23,.29,.026,1) i) What is the probability the sample proportion is between 0.27 and 0.23? i) Norm.dist(.27,.29,.026,1) - Norm.dist(.23,.29,.026,1) = 0.22-0.011 =0.21 HW7#11 Suppose 34% of the population are smartphone users and we sample 300 people. a) What is the expected sample proportion? 0.34 b) SD of the sample proportion? V: p(1-p)/n =(0.34*0.66)/300 SD: SQRTp(1-p)/n SQRT(0.34*0.66)/300 =0.027 c) If we treat the sample proportion as having a normal relative frequency, what is the distribution of the sample proportion? P~N(0.34,0.027) d) What is the probability the sample proportion is less than 0.37? Norm.dist(.37,.34,.027,1) e) What is the probability the sample proportion is greater than 0.38? 1-Norm.dist(.38,.34,.027,1) f) What is the probability the sample proportion is between 0.37 and 0.31? Norm.dist(.37,.34,.027,1) - Norm.dist(.31,.34,.027,1) g) Approximately 95% of our sample proportions are between what 2 numbers? Use the 68-95-99.7 rule OR h) Find the lower limit and the upper limit when: We create and interval beginning at 0.38 and go 2 SD in both directions? Lower: 0.38+(-2*0.027)= 0.326 Upper: 0.38+(2*0.027)= 0.434 When we create an interval starting at 0.38 and go 2 SD in both directions, is the mean inside this interval? The interval is [0.326,0.434]; The mean is 0.34 Yes. The mean 0.34 is inside the interval. P~N(0.34,0.027) i) Find the lower limit and upper limit when: We create and interval beginning at 0.41 and go 2 SD in both directions? Lower: 0.41+(-2*0.027)= 0.356 Upper: 0.41+(2*0.027)= 0.464 When we create an interval starting at 0.41 and go 2 SD in both directions, is the mean inside this interval? The interval is [0.356,0.464]; The mean is 0.34 No. The mean 0.34 is not inside the interval. j) We create and interval beginning at 0.33 and go 2 SD in both directions? Does interval contain the mean Lower: 0.33+(-2*0.027)= 0.276 Upper: 0.33+(2*0.027)= 0.384 The interval is [0.276,0.384]; The mean is 0.34 Yes. The mean 0.34 is inside the interval k) We create and interval beginning at 0.26 and go 2 SD in both directions? Does interval contain the mean Lower: 0.26+(-2*0.027)= 0.206 Upper: 0.26+(2*0.027)= 0.314 The interval is [0.206,0.314]; The mean is 0.34 No. The mean 0.34 is not inside the interval P~N(0.34,0.027) l) If I select a sample proportion at random and create a 2 SD interval, what is the probability the interval contains the population mean? Note: As long as the starting point is between the 2 SD, it will contain pop mean 95% of the time I will select a sample proportion between 0.286 and 0.394. The subsequent intervals will contain the population mean. m) If I select a sample proportion at random and create a 2 SD interval, what is the probability the interval will not contain the population mean? 1-0.95 = 0.05 5% of time will select a sample proportion less than 0.286 or more than 0.394 P~N(0.45,0.05) Mean: 0.45 SD: 0.05 1. Create a 2 SD interval starting at 0.5 Does this interval contain the population mean? a. Lower: 0.5+(-2*0.05)=0.4 b. Upper: 0.5+(2*0.05)=0.6 The interval is [0.4, 0.6]; The mean is 0.45 Yes. The mean 0.45 is inside the interval 2. Create a 2 SD interval starting at 0.57 Does this interval contain the population mean? a. Lower: 0.57+(-2*0.05)=0.47 b. Upper: 0.57+(2*0.05)=0.67 The interval is [0.47, 0.67]; The mean is 0.45 No. The mean 0.45 is not inside the interval 3. If I select a sample proportion at random and create a 2 SD interval, what is the probability the interval contains the population mean? 95% 4. If I select a sample proportion at random and create a 2 SD interval, what is the probability the interval will not contain the population mean? 1- .95= 0.05 5% P~N(0.45,0.05) Mean: 0.45 SD: 0.05 We select a sample proportion at random; and we create a 2 SD interval 1. What is the probability that the interval contains the population mean? a. 95% 2. What is the probability that the interval doesn’t contain the population mean? a. 5% 3. If we create an interval of 3 SD. What is the probability that interval contains the population mean? a. 99.7% 4. If we create an interval of 3 SD. What is the probability that interval does not contain the population mean? a. 0.3% Example We sample 300 persons and find the sample proportion is 0.4. The SD is 0.08 1. Create a 95% Confidence Interval. That is, create an interval where we are 95% sure the population mean is between those 2 numbers a. Lower: 0.4+(-2*0.08)=0.24 b. Upper: 0.4+(2*0.08)=0.56 2. Create a 99.7% Confidence Interval. a. [0.16, 0.64] 3. Create a 68% Confidence Interval. a. [0.32, 0.48] Example We sample 300 persons in the US and find the 40% of them own XBox SD: SQRT((.6*.4)/300)=0.028284271 1. What percent of people in the US own Xbox? a. We don’t know 2. Do we know the SD of people who own Xbox? a. We don’t know 3. What is the standard error. a. SQRT((.6*.4)/300)=0.028284271 4. Create a 95% confidence interval That is, create an interval where we are 95% sure the population mean is between those 2 numbers [0.344, 0.456] 5. Create a 99.7% confidence interval a. [0.316, 0.484] Excel Formula for creating Confidence Interval We sample 300 people and 40% owns game boxes. 1. Create a 93% confidence interval using the formula a. Lower: 0.4-norm.inv(0.5+9.3/2,0,1)SQRT(.4*.6/300) b. Upper: 0.4+norm.inv(0.5+9.3/2,0,1)SQRT(.4*.6/300) c. [0.35,0.45] 2. Create a 98% confidence interval using the formula a. Lower: 0.4-norm.inv(0.5+9.8/2,0,1)SQRT(.4*.6/300) b. Upper: 0.4+norm.inv(0.5+9.8/2,0,1)SQRT(.4*.6/300) c. [0.33,0.47] HW7#15 160 out of 400 students we sample own a laptop 1. Sample proportion a. 160/400= 0.4 2. Standard error of the sample proportion a. SQRT(0.4*0.6/400)=0.024 3. Construct an approximate 95% Confidence Interval using 2 SD interval [0.352,0.448] HW7#19 Formula for Sample Size n= (norm.inv(0.5+CI/2,0,1)/MOE)^2 x (p(1-p)) = (norm.inv(0.5+(.0.95/2),0,1)/0.02)^2 x (.3*.7) = 2016.77 =2017 We round up Assumptions Randomization Condition To satisfy the Randomization Condition, the sampling method has to be at least not biased and the sample has to be representative of the population. The Randomization Condition is not met because the seeds are all in the same packet; it is unlikely that they are randomized and may not be independent of each other. 10% Condition The 10% Condition states that the sample must be no larger than 10% of the population. The 10% Condition is met because seeds likely consist of less than 10% of that species seed population. Success/ Failure Condition The sample size has to be big enough so that both np and nq are at least 10 to meet the Success/Failure Condition. Calculate np and nq. ● ● The Randomization Condition and the 10% Condition are used to check the Independence Assumption. The Success/Failure Condition is used to check the Sample Size Assumption